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Anna [14]
4 years ago
11

Unpolarized light with intensity I0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal polariz

ing filter whose axis is at 42.0 ∘ to that of the first. Determine the intensity of the beam after it has passed through the second polarizer.
Physics
1 answer:
nordsb [41]4 years ago
8 0

Answer:

0.276I_0

Explanation:

When unpolarized light passes through a polarizer, only the component of the light vibrating in the direction parallel to the axis of the polarizer passes through: therefore, the intensity of light is reduced by half, since only 1 out of 2 components passes through.

So, after the first polarizer, the intensity of light passing through is:

I_1=\frac{I_0}{2}

Where I_0 is the initial intensity of the unpolarized light.

Then, the light (which is now polarized) passes through the second polarizer. Here, the intensity of the light passing through the second polarizer is given by Malus Law:

I_2=I_1 cos^2 \theta

where:

\theta is the angle between the axes of the two polarizers

In this problem the angle is

\theta=42^{\circ}

So the intensity after of light the 2nd polarizer is:

I_2=I_1 (cos 42^{\circ})^2=\frac{I_0}{2}(cos 42^{\circ})^2=0.276I_0

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7 0
3 years ago
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Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

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Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

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Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

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Plug in the given values and solve for 'μ'. This gives,

\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

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